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13 Sided Polygon of Charges

  1. Sep 7, 2012 #1
    Imagine you have a regular 13 sided polygon with charges distributed on every corner of the polygon.

    What would a test charge experience in the center?

    The answer to that was a 0 net force (which makes some intuitive sense to me due to the symmetry of the polygon). I understand that if we were to place a force in the

    However what would happen if one were to remove one of the charges?

  2. jcsd
  3. Sep 7, 2012 #2
    Just draw the 12 arrows and you´ll see the result.
  4. Sep 7, 2012 #3


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    That seems a rather tedious and time-consuming way to go about this. It is much easier to take the previous result of zero, then subtract the force that was due to the removed charge.
  5. Sep 7, 2012 #4
    Alright that seems like a reasonable way to go about this.. I guess I just want to double check too that describing the symmetry of the polygon is an adequate way to solve this problem, and if there was a more elegant way to answer the problem.
  6. Sep 8, 2012 #5


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    After thinking more, actually this is not difficult. The 12 arrows form 12 sides of a 13-sided regular polygon; the missing side gives the resulting force. Nice way to look at it!
  7. Sep 8, 2012 #6
    and a nicee idea!!
    imagine we have an N gon on xy page.we want to find the E vector at some z(out the page over the center of polygon) but not actually OVER the center.say we move away from the axis z(which starts at O which is the center of polygon) a tiny lenght delta .it means in cylindrical coordinates we are at z(z) and delta(r) and some phi(which wont matter).to the first nonzero order of delta we want the E vector.
    the idea:finding E on the z axis is fairly easy.(because of symmetry).the trick is to take a differential cylinder near the point on the axis and write divergence>>everything going in comes out.things which go in is Ez from below and Ez+dz comes out .also from the sides of cylinder E r comes out.(we want E r)equating the flux simply gives the desired E.nicee!
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